A238655 Number of partitions of n having standard deviation σ > 3.

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 5, 10, 15, 23, 33, 49, 68, 99, 133, 179, 246, 334, 432, 583, 756, 974, 1261, 1618, 2076, 2657, 3336, 4228, 5270, 6592, 8190, 10182, 12567, 15533, 19008, 23307, 28410, 34622, 42041, 50959, 61487, 74259, 88734, 106666, 127587

Offset

1,10

Comments

Regarding "standard deviation" see Comments at A238616.

Links

Table of n, a(n) for n=1..50.

Example

There are 30 partitions of 9, whose standard deviations are given by these approximations:  0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 1.

Mathematica

z = 53; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t];
s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, c[t]}]/c[t]]
Table[Count[g[n], p_ /; s[p] > 3], {n, z}]   (*A238655*)
Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st dev's of partitions of 30*)

Crossrefs

Cf. A238616, A238661, A238656, A238657.

Sequence in context: A349803 A081240 A184443 * A132295 A086651 A074495

Adjacent sequences: A238652 A238653 A238654 * A238656 A238657 A238658

Keyword

nonn,easy

Author

Clark Kimberling, Mar 03 2014

Status

approved